GS Charging CONTEST

[posplayr]

OK I have decided to have an Educational GS Charging Contest, however it is not completely educational because I'm offering to deliver this Brand Spanking New "Duanage HONDA R/R" to the winner. Duane hand picks these babies, connectorizes for easy install and tests them. If it is busted (or not) he even takes them back.

The Rules:

RULE #1.) The prize is shown below. delivered to the winner via USPS.




RULE #2) First person to correctly answer the following question by posting the CORRECT answer and the CORRECT explanation to this thread.

Determining if an answer is CORRECT should be unambiguous; it is #1,#2,#3,#4,#5, or #6.

On the other hand, you need to explain in at least simple term why the answer is correct. I'll be the judge of what is a suitable explanation.

RULE #3)
The question is:

Assuming your GS is charging correctly while you are running down the road, then your charging system is producing current to both:
a.) Charge your battery
and
b.) Provide power to the GS loads (light, coils, igniter, bells, whistles, etc).

The power originates in the stator as AC voltage and current, but the R/R transforms this power into DC power at nominally +14V.

The power and current comes from the R/R out of the R/R(+) RED leg and splits part going to the battery and part going to the GS load.

The Question is : Of the 6 option shown below which one is CORRECT? Note the Red Currents are shown correctly, so which return current figure is CORRECT?

Rule #4) Chef1366 already knows the answer and is disqualified.

Rule #5) Contest ends 6:00 PM Sunday Nite April 18th 2010



























P.S. HINT : The Suzuki Manual has the answer under "Charging System"

[kdo58]

#4 all the current comes through rr, and #4 has single ground point which works the best.

[rustybronco]

I'll bite.

#5

all (unused) currents go back to the power source, hence in your drawing it would be the source of rectification. (R/R)

[BassCliff]

Hi,

I would say #5 also. The zener in the r/r has to know when to shunt excess energy to ground when the battery is fully charged.


Thank you for your indulgence,

BassCliff

[wkmpt]

Is this a trick question? The diagram I have for my 750 shows all current flowing to ground, not towards the r/r or the battery (-) terminal.

Granted, I observe on my bike that there are several wires that connect to the battery (-), so I am going to say that return paths for the lighting, load, and r/r flow to the battery (-) terminal, which is grounded to the frame.

I choose #4.

[donimo]

I also say #5, for the same reason as BassCliff (although I am sure that reason disqualifies me haha).

as an aside, it's just "current", no "s", sorry I am a terminology whore, blame my father...

cool contest idea!

[SqDancerLynn1]

#5 also, the reg dissipates the excess current as heat

[chef1366]

Where's "all the above"?

I haven't been disqualified since war ball.

[kinnet]

I need more time! I'm in class... Stupid systems analysis and design. I'd much rather be researching this current flow matter.
Maybe I should change my major.

[Sandy]

#5 for me too. Current's burned of as heat at the R/R, that's why it needs fins.

[posplayr]

systems analysis and design. .

That sounds like an EE class

[kinnet]

It's all IT nonsense. But you could use the SDLC for EE I'm sure.
I will have an answer for this contest by the end of the night. How correct it will be is undetermined. Ha.

[Theo]

So I guess we are talking about current flowing from positive to negative rather than electron flow (negative to positive). I don't really know much about electrical theory, and am at work, so can't look at the manual, but I would guess #3. All of the current flows from the positive lead, powers all of the circuits and then makes it's final journey to the R/R and battery. Since both the battery and R/R are connected, I would think that means the current is returned to both from the GS. Even though the R/R is controlling how much voltage the whole system sees, I would think that happens with the alternator connections and not the connections that deliver +14 V. That's my 2 cents.

[chef1366]

I thought it returned to the tires.
Sorry Jim, I gave it up.

[mottyl]

OK. May be I am tired and should look at the schematic better but this is what I remember from electronics.

The heat generated by the r/r is caused by it converting ac to dc and the amount of current required to run the gs electrics. The more load (ie. headlights, dead battery etc.) on the system will require the r/r to provide more current and the r/r will have to dissapate more heat. If you disconnected every electrical load on the bike except what is required to run the engine, the current load would go down and the r/r would not run as hot.

Please ignore the first paragraph. It is relevant to power supplies just not the type they use on motorcycles.

The return path of a electrical or electronic circuit is ground. Thereticaly there should be no resistance along your ground path and therefore no current flow. This is not possible (yet) in real life. For example there should be no potential resistance between the ground at the r/r and the ground of the battery ( this would make 4 and 5 right ) . If we are tring to measure current flow along a ground then the path of most resistance would have the most current flowing through it.

Or did I screw up understanding the question?

It was a long day at work>

Chris